描述：计算逆波兰表达法的结果
Sample：
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
使用stack实现：
1 def is_op(c): 2 return c in ['+', '-', '*', '/'] 3 4 def divide(x, y): 5 if (x * y) < 0: 6 return -1 * ((-x)/y) 7 return x/y 8 9 class Solution: 10 # @param tokens, a list of string 11 # @return an integer 12 def evalRPN(self, tokens): 13 opDict = {'+': lambda x,y: x+y, 14 '-': lambda x,y: x-y, 15 '*': lambda x,y: x*y, 16 '/': divide} 17 record = [] 18 19 for item in tokens: 20 if is_op(item): 21 second = record.pop() 22 first = record.pop() 23 record.append(opDict[item](first, second)) 24 else: 25 record.append(int(item)) 26 return record[0]
使用树实现：
1 def is_op(c): 2 return c in ['+', '-', '*', '/'] 3 4 def divide(x, y): 5 if (x * y) < 0: 6 return -1 * ((-x)/y) 7 return x/y 8 9 class Tree: 10 def __init__(self, data): 11 self.data = data 12 self.parent = None 13 self.left = None 14 self.right = None 15 16 class Solution: 17 # @param tokens, a list of string 18 # @return an integer 19 def __init__(self): 20 self.opDict = {'+': lambda x,y: x+y, 21 '-': lambda x,y: x-y, 22 '*': lambda x,y: x*y, 23 '/': divide} 24 25 def builtTree(self, tokens): 26 if not is_op(tokens[-1]): 27 return int(tokens[-1]) 28 29 # if element is an operator 30 cur_tree = Tree(tokens[-1]) 31 top_tree = cur_tree 32 33 for item in tokens[-2::-1]: 34 if cur_tree.right is None: 35 if is_op(item): 36 cur_tree.right = Tree(item) 37 cur_tree.right.parent = cur_tree 38 cur_tree = cur_tree.right 39 else: 40 cur_tree.right = int(item) 41 42 if cur_tree.right and cur_tree.left: 43 cur_tree = self.getUpperNode(cur_tree) 44 continue 45 46 if cur_tree.left is None: 47 if is_op(item): 48 cur_tree.left = Tree(item) 49 cur_tree.left.parent = cur_tree 50 cur_tree = cur_tree.left 51 else: 52 cur_tree.left = int(item) 53 54 if cur_tree.right is not None and cur_tree.left is not None: 55 cur_tree = self.getUpperNode(cur_tree) 56 57 return top_tree 58 59 # Move to upper node if cur node if full. If top_node return. 60 def getUpperNode(self, node): 61 while node.right is not None and node.left is not None: 62 if node.parent is None: 63 return node 64 65 node = node.parent 66 67 return node 68 69 def getValue(self, node): 70 if type(node) is type(1): 71 return node 72 else: 73 return self.getResult(node) 74 75 def getResult(self, treeNode): 76 leftValue = self.getValue(treeNode.left) 77 rightValue = self.getValue(treeNode.right) 78 79 result = self.opDict[treeNode.data](leftValue, rightValue) 80 81 return result 82 83 def evalRPN(self, tokens): 84 topNode = self.builtTree(tokens) 85 86 if type(topNode) is type(1): 87 return topNode 88 else: 89 resultNum = self.getResult(topNode) 90 return resultNum
备注-1：if cur_node.right is None 不能用 if cur_node.right 因为cur_node.right 如果是数字0的话会有问题 当然 不转成int的话直接存string等到运算时再转int应该就可以这样写了
备注-2：python的除法跟c++不太一样 3/-5 = -1
结论：
根据问题的具体特性，选择合适的数据结构解决问题会差别很大